In the series RLC circuit the resistor has a value of R = 500.0 ohms, the capaci
ID: 1601813 • Letter: I
Question
In the series RLC circuit the resistor has a value of R = 500.0 ohms, the capacitor C = 2.00 microF, and the inductance is rated at 200.0 mH. If the generatoe frequecny f =128Hz and an rms voltage of 12.0V is applied:
a) compute the capacitive and inductive reactances (I calculated xl=161ohms, xc= 622ohms)
b) calculate the rms voltage drops through each circuit elemnt (Vr, Vc, Vl) (I calculated Vr= 8.8v, Vc=10.9v, Vl=2.8v)
C) determine the phase angle (I calculated -42.8 degrees)
d) at what frequency will the circuit resonate ( I calculated 50hz)
Explanation / Answer
Part A)
Formula
Capacitave Reactance ... XC = 1/2fC = 1/(2)(128)(2 X 10-6) = 621.7
Inductive Reactance ...XL = 2fL = (2)(128)(200 X 10-3) = 160.8
Part B)
Now we can calculate Z from Z = [(R2) + (XL - Xc)2]
Z = [(5002) + (160.8 - 621.7)2]
Z = 680
Then V = IZ
(12) = (I)(680)
I = .0176 A
Finally...
We know that formula
VR = IR = (.0176)(500) = 8.82 V
VL = IXL = (.0176)(160.8) = 2.84 V
VC = IXC = (.0176)(621.7) = 11.0 V
Part d)
Resonant Frequency ...
f = 1/2(LC)
f = 1/2(200 X 10-3)(2 X 10-6)
f = 251.6 Hz
Part c)
tan = (XL - XC)/R
tan = (160.8 - 621.7)/(500)
= -42.7o
The negative means that the current leads the applied voltage